準備

Tr[AB]=AijBji=BjiAij=Tr[BA]\begin{aligned} \text{Tr}[AB] &= A_{ij}B_{ji} \\ &= B_{ji} A_{i j} \\ &= \text{Tr}[BA] \end{aligned}

Tr[AB]=∫dψ⟨ψ∣AB∣ψ⟩=∫dψ⟨ψ∣A(∫dψ′∣ψ′⟩⟨ψ′∣)B∣ψ⟩=∫dψ∫dψ′⟨ψ∣A∣ψ′⟩⟨ψ′∣B∣ψ⟩=∫dψ′∫dψ⟨ψ′∣B∣ψ⟩⟨ψ∣A∣ψ′⟩=∫dψ′⟨ψ′∣B(∫dψ∣ψ⟩⟨ψ∣)A∣ψ′⟩=∫dψ′⟨ψ′∣BA∣ψ′⟩=Tr[BA]\begin{aligned} \text{Tr} [AB] &= \int d\psi \bra{\psi} AB \ket{\psi} \\ &= \int d\psi \bra{\psi} A \left( \int d\psi' \ket{\psi'}\bra{\psi'} \right) B \ket{\psi} \\ &= \int d\psi \int d\psi' \bra{\psi} A \ket{\psi'}\bra{\psi'} B \ket{\psi} \\ &= \int d\psi' \int d\psi \bra{\psi'} B \ket{\psi} \bra{\psi} A \ket{\psi'} \\ &= \int d\psi' \bra{\psi'} B \left( \int d\psi \ket{\psi}\bra{\psi} \right) A \ket{\psi'} \\ &= \int d\psi' \bra{\psi'} BA \ket{\psi'} \\ &= \text{Tr} [BA] \end{aligned}

⟨AB⟩eq=∑ψ⟨ψ∣ABπ∣ψ⟩=Tr[ABπ]=Tr[πAB]     (∵トレースの循環性)=Tr[BπA]     (∵トレースの循環性)≠Tr[πBA]     (ie.≠⟨BA⟩eq)\begin{aligned} \langle A B \rangle_{eq} &= \sum_{\psi} \bra{\psi} A B \pi \ket{\psi} \\ &= \text{Tr}\left[ A B \pi \right] \\ &= \text{Tr}\left[ \pi A B \right] \ \ \ \ \ (\because \text{トレースの循環性}) \\ &= \text{Tr}\left[ B \pi A \right] \ \ \ \ \ (\because \text{トレースの循環性}) \\ &\neq \text{Tr}\left[ \pi B A \right] \ \ \ \ \ (ie. \neq \langle B A \rangle_{eq}) \\ \end{aligned}

U(t)=e−ihHtU(t)†=U(t)−1=eihHt\begin{align*} &U_{(t)} = e^{-\frac{i}{h}Ht} \\ &U^{\dagger}_{(t)} = U^{-1}_{(t)} = e^{\frac{i}{h}Ht} \\ \end{align*}

A(t)=U(t)†AU(t)A˙(t)=ddt(U(t)†AU(t))=U(t)†iℏHAU(t)−U(t)†AiℏHU(t)=U(t)†iℏ[H,A]U(t)=U(t)†A˙U(t)\begin{aligned} A_{(t)} &= U^{\dagger}_{(t)} A U_{(t)} \\ \dot{A}_{(t)} &= \frac{d}{dt} \left( U^{\dagger}_{(t)} A U_{(t)} \right) \\ &= U^{\dagger}_{(t)} \frac{i}{\hbar} H A U_{(t)} - U^{\dagger}_{(t)} A \frac{i}{\hbar} H U_{(t)} \\ &= U^{\dagger}_{(t)} \frac{i}{\hbar} [H, A] U_{(t)} \\ &= U^{\dagger}_{(t)} \dot{A} U_{(t)} \end{aligned}

iℏ∂∂t∣ψ(t)⟩=H(t)∣ψ(t)⟩∴iℏ∂∂tU(t)=H(t)U(t)∴U(t+Δt)=(I−iℏH(t)Δt)U(t)\begin{aligned} i \hbar \frac{\partial}{\partial t} \ket{\psi_{(t)}} &= H_{(t)} \ket{\psi_{(t)}} \\ \therefore i \hbar \frac{\partial}{\partial t} U_{(t)} &= H_{(t)} U_{(t)} \\ \therefore U_{(t + \Delta t)} &= \left( I - \frac{i}{\hbar} H_{(t)} \Delta t \right) U_{(t)} \end{aligned}

逐次積分

U(Δt)=(I−iℏH(0)Δt)U(0)=(I−iℏH(0)Δt)U(2Δt)=(I−iℏH(Δt)Δt)U(Δt)=(I−iℏH(Δt)Δt)(I−iℏH(0)Δt)⋯U(nΔt)=∏in(I−iℏH(iΔt)Δt)∴U(τ)=lim⁡n→∞∏in(I−iℏH(iτn)τn)=I−iℏ∫0τdtH(t)+(iℏ)2∫τ′τdt∫0τ′dt′H(dt)H(dt′)+⋯≔exp⁡←(−iℏ∫0τHdt)     (時間順序指数)\begin{aligned} U_{(\Delta t)} &= \left(I - \frac{i}{\hbar}H_{(0)} \Delta t \right) U_{(0)} = \left( I - \frac{i}{\hbar} H_{(0)} \Delta t \right) \\ U_{(2\Delta t)} &= \left(I - \frac{i}{\hbar}H_{(\Delta t)} \Delta t \right) U_{(\Delta t)} = \left( I - \frac{i}{\hbar} H_{(\Delta t)} \Delta t \right) \left( I - \frac{i}{\hbar} H_{(0)} \Delta t \right) \\ \cdots \\ U_{(n\Delta t)} &= \prod^{n}_{i} \left( I - \frac{i}{\hbar} H_{(i \Delta t)} \Delta t \right) \\ \therefore U_{(\tau)} &= \lim_{n \to \infty} \prod^{n}_{i} \left( I - \frac{i}{\hbar} H_{(\frac{i\tau}{n})} \frac{\tau}{n} \right) = I - \frac{i}{\hbar} \int^{\tau}_0 dt H_{(t)} + \left( \frac{i}{\hbar} \right)^2 \int^{\tau}_{\tau'} dt \int^{\tau'}_0 dt' H_{(dt)} H_{(dt')} + \cdots \\ &\coloneqq \exp_{\leftarrow} \left( -\frac{i}{\hbar} \int^{\tau}_0 H dt \right) \ \ \ \ \ (\text{時間順序指数}) \end{aligned}

古典系

⟨A(t)⟩eq≔⟨eL0†tA⟩eq=∫dΓ(eL0†tA)π=∫dΓAeL0tπ=∫dΓAπ(−t)=∫dΓAπ=⟨A⟩eq\begin{aligned} \langle A_{(t)} \rangle_{eq} &\coloneqq \langle e^{L^{\dagger}_0 t} A \rangle_{eq} \\ &= \int d\Gamma (e^{L^{\dagger}_0 t} A) \pi \\ &= \int d\Gamma A e^{L_0 t} \pi \\ &= \int d\Gamma A \pi_{(-t)} \\ &= \int d\Gamma A \pi \\ &= \langle A \rangle_{eq} \end{aligned}

量子系

⟨P−1AP⟩eq=Tr[P−1AP]=Tr[APP−1]=Tr[A]=⟨A⟩eq\begin{aligned} \langle P^{-1} A P \rangle_{eq} &= \text{Tr} [P^{-1} A P] \\ &= \text{Tr} [A P P^{-1}] \\ &= \text{Tr}[A] \\ &= \langle A \rangle_{eq} \end{aligned}

⟨A(t)⟩eq=⟨Ut†AUt⟩eq=⟨A⟩eq     (∵Utがユニタリ)\langle A_{(t)} \rangle_{eq} = \langle U^{\dagger}_t A U_t \rangle_{eq} = \langle A \rangle_{eq} \ \ \ \ \ (\because U_t\text{がユニタリ})

⟨A(t)B(τ)⟩eq=⟨Ut†AUtUτ†BUτ⟩eq=⟨AUtUτ†B(τ−t)UτUt†⟩eq=⟨AU(τ−t)†B(τ−t)U(τ−t)⟩eq=⟨AB(τ−t)⟩eq\begin{aligned} \langle A_{(t)} B_{(\tau)} \rangle_{eq} &= \langle U^{\dagger}_t A U_t U^{\dagger}_\tau B U_\tau \rangle_{eq} \\ &= \langle A U_t U^{\dagger}_\tau B_{(\tau - t)} U_\tau U^\dagger_t \rangle_{eq} \\ &= \langle A U^{\dagger}_{(\tau - t)} B_{(\tau - t)} U_{(\tau - t)} \rangle_{eq} \\ &= \langle A B_{(\tau - t)} \rangle_{eq} \end{aligned}

⟨B˙(τ)A(t)⟩eq=⟨(ddτB(τ))A(t)⟩eq=ddτ⟨B(τ)A(t)⟩eq=ddτ⟨BA(t−τ)⟩eq=−ddt⟨BA(t−τ)⟩eq=−⟨BA˙(t−τ)⟩eq\begin{aligned} \left\langle \dot{B}_{(\tau)} A_{(t)} \right\rangle_{eq} &= \left\langle \left( \frac{d}{d\tau} B_{(\tau)} \right) A_{(t)} \right\rangle_{eq} \\ &= \frac{d}{d\tau} \left\langle B_{(\tau)} A_{(t)} \right\rangle_{eq} \\ &= \frac{d}{d\tau} \left\langle B A_{(t - \tau)} \right\rangle_{eq} \\ &= - \frac{d}{dt} \left\langle B A_{(t - \tau)} \right\rangle_{eq} \\ &= - \left\langle B \dot{A}_{(t - \tau)} \right\rangle_{eq} \\ \end{aligned}

2.2 古典粒子系に対する揺動散逸定理 (続き)

応答関数を計算するときは、系が平衡状態(H=Htot)(H = H_{tot})のときを考えればいよい。

α+O(ϵ)≔ϕx˙x(t;Htot−ϵxF(t))α=ϕx˙x(t;Htot−ϵxF(t))∣ϵ=0=ϕx˙x(t;Htot)\begin{aligned} \alpha + \text{O}_{(\epsilon)} \coloneqq \phi_{\dot{x}x(t; H_{tot} - \epsilon x F_{(t)})} \\ \alpha = \left. \phi_{\dot{x}x(t; H_{tot} - \epsilon x F_{(t)})} \right|_{\epsilon = 0} = \phi_{\dot{x}x(t; H_{tot})} \end{aligned}

⟨x˙(τ)⟩≃ϵ∫−∞τdtF(t)ϕx˙x(τ−t;Htot−ϵxF(t))=ϵ∫−∞τdtF(t)(α+O(ϵ))≃ϵ∫−∞τdtF(t)α=ϵ∫−∞τdtF(t)ϕx˙x(τ−t;Htot)\begin{aligned} \langle \dot{x}_{(\tau)} \rangle &\simeq \epsilon \int^{\tau}_{-\infty} dt F_{(t)} \phi_{\dot{x}x(\tau - t; H_{tot} - \epsilon x F_{(t)})} \\ &= \epsilon \int^{\tau}_{-\infty} dt F_{(t)} \left( \alpha + \text{O}_{(\epsilon)} \right) \\ &\simeq \epsilon \int^{\tau}_{-\infty} dt F_{(t)} \alpha \\ &= \epsilon \int^{\tau}_{-\infty} dt F_{(t)} \phi_{\dot{x}x(\tau - t; H_{tot})} \end{aligned}

応答関数 ϕx˙x\phi_{\dot{x}x}

x˙(t)=eL0†tx˙=eL0†tL0†x=ddteL0†tx=ddtx(t)\begin{aligned} \dot{x}_{(t)} &= e^{L^{\dagger}_0 t} \dot{x} \\ &= e^{L^{\dagger}_0 t} L^{\dagger}_0 x \\ &= \frac{d}{dt} e^{L^{\dagger}_0 t} x \\ &= \frac{d}{dt} x_{(t)} \end{aligned}

ϕx˙x(τ−t)≔β⟨x˙(τ−t)x˙⟩eq=β⟨x˙(τ)x˙(t)⟩eq=β⟨x˙(τ)ddtx(t)⟩eq=βddt⟨x˙(τ)x(t)⟩eq\begin{aligned} \phi_{\dot{x}x(\tau - t)} &\coloneqq \beta \langle \dot{x}_{(\tau - t)} \dot{x} \rangle_{eq} \\ &= \beta \langle \dot{x}_{(\tau)} \dot{x}_{(t)} \rangle_{eq} \\ &= \beta \langle \dot{x}_{(\tau)} \frac{d}{dt} x_{(t)} \rangle_{eq} \\ &= \beta \frac{d}{dt} \langle \dot{x}_{(\tau)} x_{(t)} \rangle_{eq} \end{aligned}

長時間後に一定になった速度 ⋯(2.11)\cdots (2.11)

⟨x˙(τ)⟩F0=∫−∞τdtϕx˙x(τ−t)=∫0∞dtϕx˙x(t)     (t⇒τ−t)=β∫0∞dt⟨x˙(t)x˙⟩eq∴⟨x˙(τ)⟩eqF0=⟨x˙⟩eqF0=β∫0∞dt⟨x˙(t)x˙⟩eq\begin{aligned} \frac{\langle \dot{x}_{(\tau)} \rangle}{F_0} &= \int^{\tau}_{-\infty} dt \phi_{\dot{x}x(\tau - t)} \\ &= \int^{\infty}_0 dt \phi_{\dot{x}x(t)} \ \ \ \ \ (t \Rightarrow \tau - t) \\ &= \beta \int^{\infty}_0 dt \langle \dot{x}_{(t)} \dot{x} \rangle_{eq} \\ \therefore \frac{\langle \dot{x}_{(\tau)} \rangle_{eq}}{F_0} &= \frac{\langle \dot{x} \rangle_{eq}}{F_0} = \beta \int^{\infty}_0 dt \langle \dot{x}_{(t)} \dot{x} \rangle_{eq} \end{aligned}

ランジュバン方程式でのμ\mu

mx¨=F−γx˙+ξm \ddot{x} = F - \gamma \dot{x} + \xi

m⟨x¨⟩=F−γ⟨x˙⟩+⟨ξ⟩∴F=γ⟨x˙⟩     (∵速度一定)∴⟨x˙⟩F=μ=12kBT∫−∞∞ds⟨x˙(s)x˙⟩     (ref. (1.12))\begin{aligned} m \langle \ddot{x} \rangle &= F - \gamma \langle \dot{x} \rangle + \langle \xi \rangle \\ \therefore F &= \gamma \langle \dot{x} \rangle \ \ \ \ \ (\because \text{速度一定}) \\ \therefore \frac{\langle \dot{x} \rangle}{F} &= \mu = \frac{1}{2k_B T} \int^{\infty}_{-\infty} ds \langle \dot{x}_{(s)} \dot{x} \rangle \ \ \ \ \ (\text{ref. } (1.12)) \end{aligned}

パルス入力に対する応答 ⋯(2.12)\cdots (2.12)

F(t)=F0δ(t−t′)F_{(t)} = F_0 \delta_{(t - t')}

⟨x˙(τ)⟩=ϵ∫−∞τdtF(t)ϕxx˙(τ−t)=ϵ∫−∞τdtF0δ(t−t′)ϕxx˙(τ−t)=ϵF(0)ϕxx˙(τ−t′)     (τ≥t′)\begin{aligned} \langle \dot{x}_{(\tau)} \rangle &= \epsilon \int^{\tau}_{-\infty} dt F_{(t)} \phi_{x\dot{x}(\tau - t)} \\ &= \epsilon \int^{\tau}_{-\infty} dt F_0 \delta_{(t - t')} \phi_{x\dot{x}(\tau - t)} \\ &= \epsilon F_{(0)} \phi_{x\dot{x}(\tau - t')} \ \ \ \ \ (\tau \ge t') \end{aligned}

フーリエ表示の揺動散逸定理 ⋯(2.15)\cdots (2.15)

χx˙x(ω)=χ(ω)′+iχ(ω)′′\chi_{\dot{x}x(\omega)} = \chi'_{(\omega)} + i \chi''_{(\omega)}

χ(ω)′=Re[∫0∞dtβ⟨x˙(t)x˙⟩eqeiωt]=β∫0∞dtC(t)Re[eiωt]=β∫0∞dtC(t)cos⁡(ωt)=β2∫0∞dtC(t)(eiωt+e−iωt)=β2(∫0∞dtC(t)eiωt−∫0∞(−dt)C(−t)e−iωt)=β2(∫0∞dtC(t)eiωt+∫−∞0dtC(t)eiωt)=β2C(ω)\begin{aligned} \chi'_{(\omega)} &= \text{Re} \left[ \int^{\infty}_{0} dt \beta \langle \dot{x}_{(t)} \dot{x} \rangle_{eq} e^{i\omega t} \right] \\ &= \beta \int^{\infty}_{0} dt C_{(t)} \text{Re} [e^{i\omega t}] \\ &= \beta \int^{\infty}_0 dt C_{(t)} \cos{(\omega t)} \\ &= \frac{\beta}{2} \int^{\infty}_{0} dt C_{(t)} \left( e^{i \omega t} + e^{-i \omega t} \right) \\ &= \frac{\beta}{2} \left( \int^{\infty}_{0} dt C_{(t)} e^{i \omega t} - \int^{\infty}_0 (-dt) C_{(-t)} e^{-i \omega t} \right) \\ &= \frac{\beta}{2} \left( \int^{\infty}_{0} dt C_{(t)} e^{i \omega t} + \int^0_{-\infty} dt C_{(t)} e^{i \omega t} \right) \\ &= \frac{\beta}{2} C_{(\omega)} \end{aligned}

2.2 線形応答理論

§ 2.2.1 力学応答の線形応答

Htot(t)=H−BF(t)H_{tot(t)} = H - BF_{(t)}

タイプ 1

A∝BA \propto B

タイプ 2

A∝B˙A \propto \dot{B}

§ 2.2.1.1 一般論

時間発展演算子 ⋯(2.18)\cdots (2.18)

iℏ∂∂tU(t)=[H−FB]U(t)Uˉ(t)≔eiℏHtU(t)\begin{aligned} i \hbar \frac{\partial}{\partial t} U_{(t)} &= [H - FB] U_{(t)} \\ \bar{U}_{(t)} &\coloneqq e^{\frac{i}{\hbar}Ht} U_{(t)} \\ \end{aligned}

∂∂tUˉ(t)=iℏHUˉ(t)−iℏeiℏHt[H−F(t)B]U(t)=iℏeiℏHtF(t)BU(t)=iℏF(t)(eiℏHtBe−iℏHt)eiℏHtU(t)=F(t)B(t)Uˉ(t)\begin{align*} \frac{\partial}{\partial t} \bar{U}_{(t)} &= \frac{i}{\hbar} H \bar{U}_{(t)} - \frac{i}{\hbar} e^{\frac{i}{\hbar}Ht}[H - F_{(t)}B]U_{(t)} \\ &= \frac{i}{\hbar} e^{\frac{i}{\hbar}Ht} F_{(t)}B U_{(t)} \\ &= \frac{i}{\hbar} F_{(t)} \left( e^{\frac{i}{\hbar}Ht} B e^{-\frac{i}{\hbar}Ht} \right) e^{\frac{i}{\hbar} Ht} U_{(t)} \\ &= F_{(t)}B_{(t)} \bar{U}_{(t)} \\ \end{align*}

[note] FFについて一次の近似を考えるので、B の時間発展を考える時は、FFを考慮せず平衡状態のハミルトニアンで時間発展すると考えて良い。

Uˉ(t+Δt)=(I+iℏF(t)B(t)Δt)Uˉ(t)∴U(t+Δt)=e−iℏH(t+Δt)(I+iℏF(t)B(t)Δt)eiℏHtU(t)∴U(t)=e−iℏHt∏in(e−iℏHΔt(I+iℏF(iΔt)B(iΔt)Δt))eiℏHt0→e−iℏHtexp⁡←(iℏ∫t0tdt′F(t′)B(t′))eiℏHt0     (Δt→0)\begin{aligned} \bar{U}_{(t + \Delta t)} &= \left( I + \frac{i}{\hbar} F_{(t)}B_{(t)} \Delta t \right) \bar{U}_{(t)} \\ \therefore U_{(t + \Delta t)} &= e^{- \frac{i}{\hbar}H(t+\Delta t)} \left( I + \frac{i}{\hbar} F_{(t)}B_{(t)} \Delta t \right) e^{\frac{i}{\hbar}Ht} U_{(t)} \\ \therefore U_{(t)} &= e^{- \frac{i}{\hbar} H t} \prod^{n}_{i} \left( e^{- \frac{i}{\hbar} H \Delta t} \left( I + \frac{i}{\hbar} F_{(i\Delta t)}B_{(i \Delta t)} \Delta t \right) \right) e^{\frac{i}{\hbar}Ht_0} \\ &\to e^{- \frac{i}{\hbar} Ht} \exp_{\leftarrow} \left( \frac{i}{\hbar} \int^t_{t_0} dt' F_{(t')} B_{(t')} \right) e^{\frac{i}{\hbar}Ht_0} \ \ \ \ \ (\Delta t \to 0) \end{aligned}

U(t)=e−iℏHt∏in(e−iℏHΔt(I+iℏF(iΔt)B(iΔt)Δt))eiℏHt0=e−iℏHt(e−iℏHΔt)n[1+∑in(iℏF(iΔt)B(iΔt)Δt)]eiℏHt0+O(F2)→e−iℏHt(1+iℏ∫t0tdt′F(t′)B(t′))eiℏHt0+O(F2)\begin{aligned} U_{(t)} &= e^{- \frac{i}{\hbar} H t} \prod^{n}_{i} \left( e^{- \frac{i}{\hbar} H \Delta t} \left( I + \frac{i}{\hbar} F_{(i\Delta t)}B_{(i \Delta t)} \Delta t \right) \right) e^{\frac{i}{\hbar}Ht_0} \\ &= e^{- \frac{i}{\hbar} H t} \left( e^{- \frac{i}{\hbar} H \Delta t} \right)^n \left[ 1 + \sum^{n}_{i} \left( \frac{i}{\hbar} F_{(i\Delta t)}B_{(i \Delta t)} \Delta t \right) \right] e^{\frac{i}{\hbar}Ht_0} + \text{O}(F^2) \\ &\to e^{- \frac{i}{\hbar} H t} \left( 1 + \frac{i}{\hbar} \int^t_{t_0} dt' F_{(t')} B_{(t')} \right) e^{\frac{i}{\hbar} Ht_0} + \text{O}(F^2) \end{aligned}

演算子の期待値の時間発展 ⋯(2.19)\cdots (2.19)

A(τ)=U(τ)†AU(τ)=e−iℏHt0(1−iℏ∫t0τdt′F(t′)B(t′))eiℏHτAe−iℏHτ(1+iℏ∫t0τdt′F(t′)B(t′))eiℏHt0+O(F2)=eiℏH(τ−t0)Ae−iℏH(τ−t0)+e−iℏHt0[eiℏHτAe−iℏHτ,iℏ∫t0τdt′F(t′)B(t′)]eiℏHt0+O(F2)=eiℏH(τ−t0)Ae−iℏH(τ−t0)+iℏ∫t0τdt′F(t′)[A(τ−t0),B(t′−t0)]+O(F2)\begin{aligned} A_{(\tau)} &= U^{\dagger}_{(\tau)} A U_{(\tau)} \\ &= e^{- \frac{i}{\hbar} H t_0} \left( 1 - \frac{i}{\hbar} \int^\tau_{t_0} dt' F_{(t')} B_{(t')} \right) e^{\frac{i}{\hbar} H \tau} A e^{- \frac{i}{\hbar} H \tau} \left( 1 + \frac{i}{\hbar} \int^\tau_{t_0} dt' F_{(t')} B_{(t')} \right) e^{\frac{i}{\hbar} Ht_0} + \text{O}(F^2) \\ &= e^{\frac{i}{\hbar} H (\tau - t_0)} A e^{-\frac{i}{\hbar} H (\tau - t_0)} + e^{-\frac{i}{\hbar}Ht_0} \left[ e^{\frac{i}{\hbar} H \tau} A e^{- \frac{i}{\hbar} H \tau}, \frac{i}{\hbar} \int^\tau_{t_0} dt' F_{(t')}B_{(t')} \right] e^{\frac{i}{\hbar}Ht_0} + \text{O}(F^2) \\ &= e^{\frac{i}{\hbar} H (\tau - t_0)} A e^{-\frac{i}{\hbar} H (\tau - t_0)} + \frac{i}{\hbar} \int^\tau_{t_0} dt' F_{(t')} [ A_{(\tau - t_0)}, B_{(t' - t_0)} ] + \text{O}(F^2) \\ \end{aligned}

[note] FFの一次近似を考えているので、積分の中では非平衡の時間発展と平衡の時間発展を区別する必要がない。

⟨A(τ)⟩eq−⟨A⟩eq=⟨eiℏH(τ−t0)Ae−iℏH(τ−t0)⟩eq+iℏ∫t0τdt′F(t′)⟨[A(τ−t0),B(t′−t0)]⟩eq−⟨A⟩eq+O(F2)=⟨A⟩eq+iℏ∫t0τdt′F(t′)⟨[A(τ−t′),B]⟩eq−⟨A⟩eq+O(F2)=iℏ∫t0τdt′F(t′)⟨[A(τ−t′),B]⟩eq+O(F2)\begin{aligned} \langle A_{(\tau)} \rangle_{eq} - \langle A \rangle_{eq} &= \langle e^{\frac{i}{\hbar} H (\tau - t_0)} A e^{-\frac{i}{\hbar} H (\tau - t_0)} \rangle_{eq} + \frac{i}{\hbar} \int^\tau_{t_0} dt' F_{(t')} \langle [ A_{(\tau - t_0)}, B_{(t' - t_0)} ] \rangle_{eq} - \langle A \rangle_{eq} + \text{O}(F^2) \\ &= \langle A \rangle_{eq} + \frac{i}{\hbar} \int^\tau_{t_0} dt' F_{(t')} \langle [ A_{(\tau - t')}, B ] \rangle_{eq} - \langle A \rangle_{eq} + \text{O}(F^2) \\ &= \frac{i}{\hbar} \int^\tau_{t_0} dt' F_{(t')} \langle [ A_{(\tau - t')}, B ] \rangle_{eq} + \text{O}(F^2) \\ \end{aligned}

十分に時間が経った時の期待値の変化量 ⋯(2.20)\cdots (2.20)

⟨ΔA(τ)⟩eq≔lim⁡t0→−∞⟨A(τ)⟩eq−⟨A⟩eq=iℏ∫−∞τdtF(t)⟨[A(τ−t),B]⟩eq=iℏ∫0∞dtF(τ−t)⟨[A(t),B]⟩eq     (t→τ−t)=∫0∞dtϕAB(t)F(τ−t)\begin{aligned} \langle \Delta A_{(\tau)} \rangle_{eq} &\coloneqq \lim_{t_0 \to - \infty} \langle A_{(\tau)} \rangle_{eq} - \langle A \rangle_{eq} \\ &= \frac{i}{\hbar} \int^{\tau}_{- \infty} dt F_{(t)} \langle [ A_{(\tau - t)}, B] \rangle_{eq} \\ &= \frac{i}{\hbar} \int^{\infty}_{0} dt F_{(\tau - t)} \langle [ A_{(t)}, B] \rangle_{eq} \ \ \ \ \ (t \to \tau - t) \\ &= \int^{\infty}_{0} dt \phi_{AB(t)} F_{(\tau - t)} \end{aligned}

量子系の応答関数 ⋯(2.21)\cdots (2.21)

ϕAB(t)≔iℏ⟨[A(t),B]⟩eq\phi_{AB(t)} \coloneqq \frac{i}{\hbar} \langle [A_{(t)}, B] \rangle_{eq}

⟨ΔA(τ)⟩eq=F0ϕAB(τ)     (when. F=F0δ(t))\begin{aligned} \langle \Delta A_{(\tau)} \rangle_{eq} = F_0 \phi_{AB(\tau)} \ \ \ \ \ (\text{when. }F = F_0 \delta_{(t)}) \\ \end{aligned}

ϕAB(t)=iℏ⟨[A(t),B]⟩eq=iℏTr[π[A(t),B]]=iℏ(Tr[πA(t)B]−Tr[πBA(t)])=−1iℏ(−Tr[A(t)Bπ]+Tr[A(t)πB])=1iℏTr[A(t)[π,B]]\begin{aligned} \phi_{AB(t)} &= \frac{i}{\hbar}\langle [A_{(t)}, B] \rangle_{eq} \\ &= \frac{i}{\hbar} \text{Tr} \left[ \pi [A_{(t)}, B] \right] \\ &= \frac{i}{\hbar} \left( \text{Tr} \left[ \pi A_{(t)} B \right] - \text{Tr} \left[ \pi B A_{(t)} \right] \right) \\ &= - \frac{1}{i\hbar} \left( - \text{Tr} \left[ A_{(t)} B \pi \right] + \text{Tr} \left[ A_{(t)} \pi B \right] \right) \\ &= \frac{1}{i\hbar} \text{Tr} \left[ A_{(t)} [\pi, B] \right] \end{aligned}

カノニカル相関 ⋯(2.22)∼(2.27)\cdots (2.22) \sim (2.27)

⟨X;Y⟩≔1β∫0βduTr[euHXe−uHYπ]=1β∫0βdu(Tr[euHXe−uHYπ])†=1β∫0βduTr[πYe−uHXeuH]=1β∫0βduTr[euHπYe−uHX]=1β∫0βduTr[πeuHYe−uHX]=⟨Y;X⟩\begin{aligned} \langle X; Y \rangle &\coloneqq \frac{1}{\beta} \int^{\beta}_0 du \text{Tr} \left[ e^{uH} X e^{-u H} Y \pi \right] \\ &= \frac{1}{\beta} \int^{\beta}_0 du \left( \text{Tr} \left[ e^{uH} X e^{-u H} Y \pi \right] \right)^{\dagger} \\ &= \frac{1}{\beta} \int^{\beta}_0 du \text{Tr} \left[ \pi Y e^{-u H} X e^{uH} \right] \\ &= \frac{1}{\beta} \int^{\beta}_0 du \text{Tr} \left[ e^{uH} \pi Y e^{-u H} X \right] \\ &= \frac{1}{\beta} \int^{\beta}_0 du \text{Tr} \left[ \pi e^{uH} Y e^{-u H} X \right] \\ &= \langle Y; X \rangle \end{aligned}

⟨X;Y⟩=1β∫0βdu⟨X(−iℏu)Y⟩eq→⟨XY⟩eq     (ℏ→0)\langle X; Y \rangle = \frac{1}{\beta} \int^{\beta}_0 du \langle X_{(-i\hbar u)} Y \rangle_{eq} \to \langle X Y \rangle_{eq} \ \ \ \ \ (\hbar \to 0)

[π,B]=1Z[e−βH,B]=1Z(e−βHB−Be−βH)=1Z(e−βHBeβH−B)e−βH=1Z[e−uHBeuH]u=0u=βe−βH=1Z∫0βduddu(e−uHBeuH)e−βH=iℏZ∫0βdudd(iℏu)(eiℏH(iℏu)Be−iℏH(iℏu))e−βH=iℏZ∫0βduB˙(iℏu)e−βH=iℏ∫0βduB˙(iℏu)π[π,B]=1Ze−βH(B−eβHBe−βH)=⋯=iℏπ∫0βduB˙(−iℏu)\begin{aligned} [\pi, B] &= \frac{1}{Z} [e^{-\beta H}, B] \\ &= \frac{1}{Z} (e^{-\beta H} B - B e^{-\beta H}) \\ &= \frac{1}{Z} (e^{-\beta H} B e^{\beta H} - B) e^{-\beta H} \\ &= \frac{1}{Z} \left[ e^{-u H} B e^{u H} \right]^{u = \beta}_{u = 0} e^{-\beta H} \\ &= \frac{1}{Z} \int^{\beta}_0 du \frac{d}{du} (e^{-u H} B e^{u H}) e^{-\beta H} \\ &= \frac{i\hbar}{Z} \int^{\beta}_0 du \frac{d}{d(i \hbar u)} (e^{\frac{i}{\hbar} H (i \hbar u)} B e^{- \frac{i}{\hbar} H (i \hbar u)}) e^{-\beta H} \\ &= \frac{i\hbar}{Z} \int^{\beta}_0 du \dot{B}_{(i\hbar u)} e^{-\beta H} \\ &= i\hbar \int^{\beta}_0 du \dot{B}_{(i\hbar u)} \pi \\ \\ [\pi, B] &= \frac{1}{Z} e^{-\beta H} (B - e^{\beta H} B e^{-\beta H}) \\ &= \cdots \\ &= i\hbar \pi \int^{\beta}_0 du \dot{B}_{(-i\hbar u)} \end{aligned}

ϕAB(t)=1iℏTr[A(t)[π,B]]=1iℏTr[A(t)iℏ∫0βduB˙(iℏu)π]\begin{aligned} \phi_{AB(t)} &= \frac{1}{i\hbar} \text{Tr} \left[A_{(t)} [\pi, B] \right] \\ &= \frac{1}{i\hbar} \text{Tr} \left[ A_{(t)} i\hbar \int^{\beta}_0 du \dot{B}_{(i\hbar u)} \pi \right] \\ \end{aligned}

ϕAB(t)=1iℏTr[A(t)[π,B]]=1iℏTr[A(t)iℏ∫0βduB˙(iℏu)π]=∫0βduTr[A(t)B˙(iℏu)π]=∫0βdu⟨A(t)B˙(iℏu)⟩eq=∫0βdu⟨A(t−iℏu)B˙⟩eq=β⟨A(t);B˙⟩=β⟨B˙;A(t)⟩\begin{aligned} \phi_{AB(t)} &= \frac{1}{i\hbar} \text{Tr} \left[A_{(t)} [\pi, B] \right] \\ &= \frac{1}{i\hbar} \text{Tr} \left[ A_{(t)} i\hbar \int^{\beta}_0 du \dot{B}_{(i\hbar u)} \pi \right] \\ &= \int^\beta_0 du \text{Tr} \left[ A_{(t)} \dot{B}_{(i\hbar u)} \pi \right] \\ &= \int^\beta_0 du \langle A_{(t)} \dot{B}_{(i\hbar u)} \rangle_{eq} \\ &= \int^\beta_0 du \langle A_{(t - i\hbar u)} \dot{B} \rangle_{eq} \\ &= \beta \langle A_{(t)}; \dot{B} \rangle \\ &= \beta \langle \dot{B}; A_{(t)} \rangle \end{aligned}

ϕAB(t)→β⟨A(t)B˙⟩eq     (ℏ→0)\phi_{AB(t)} \to \beta \langle A_{(t)} \dot{B} \rangle_{eq} \ \ \ \ \ (\hbar \to 0)

複素アドミッタンス ⋯(2.28)\cdots (2.28)

χAB(ω)≔∫0∞dtϕAB(t)eiωt=χ′+iχ′′\chi_{AB(\omega)} \coloneqq \int^{\infty}_0 dt \phi_{AB(t)} e^{i\omega t} = \chi' + i \chi''

χAB(−ω)=∫0∞dtϕAB(t)e−iωt=∫0∞dtϕAB(t)(eiωt)∗=(∫0∞dtϕAB(t)eiωt)∗=χAB(ω)∗\begin{aligned} \chi_{AB(-\omega)} &= \int^{\infty}_0 dt \phi_{AB(t)} e^{-i \omega t} \\ &= \int^{\infty}_0 dt \phi_{AB(t)} \left( e^{i \omega t} \right)^{*} \\ &= \left( \int^{\infty}_0 dt \phi_{AB(t)} e^{i \omega t} \right)^{*} \\ &= \chi_{AB(\omega)}^{*} \end{aligned}

演算子の応答 ⋯(2.29)\cdots (2.29)

F(t)=F0cos⁡(ωt)F_{(t)} = F_0 \cos{(\omega t)}

⟨ΔA(τ)⟩eq=∫0∞dtϕAB(t)F(τ−t)=∫0∞dtϕAB(t)F0cos⁡ω(τ−t)=F0∫0∞dtϕAB(t)Re[eiω(τ−t)]=F0Re[∫0∞dtϕAB(t)eiω(τ−t)]=F0Re[∫0∞dtϕAB(t)e−iωteiωτ]=F0Re[χAB(ω)∗eiωτ]=F0Re[χAB(ω)e−iωτ]\begin{aligned} \langle \Delta A_{(\tau)} \rangle_{eq} &= \int^{\infty}_{0} dt \phi_{AB(t)} F_{(\tau - t)} \\ &= \int^{\infty}_{0} dt \phi_{AB(t)} F_0 \cos{\omega (\tau - t)} \\ &= F_0 \int^{\infty}_{0} dt \phi_{AB(t)} \text{Re} [e^{i \omega (\tau - t)}] \\ &= F_0 \text{Re} \left[ \int^{\infty}_{0} dt \phi_{AB(t)} e^{i \omega (\tau - t)} \right] \\ &= F_0 \text{Re} \left[ \int^{\infty}_{0} dt \phi_{AB(t)} e^{- i \omega t} e^{i \omega \tau} \right] \\ &= F_0 \text{Re} \left[ \chi_{AB(\omega)}^{*} e^{i \omega \tau} \right] \\ &= F_0 \text{Re} \left[ \chi_{AB(\omega)} e^{-i \omega \tau} \right] \\ \end{aligned}

§ 2.2.1.2 クラマース・クローニッヒ関係

リーマン・ルベーグの補題

lim⁡ω→0∣∫0∞dtϕAB(t)eiωt∣=0     (Im[ω]>0)\lim_{\omega \to 0} \left| \int^{\infty}_0 dt \phi_{AB(t)} e^{i \omega t} \right| = 0 \ \ \ \ \ (\text{Im}[\omega] > 0)

I≔∫0∞dtϕAB(t)eiωt=∫−πω∞dsϕAB(s+πω)eiω(s+πω)     (s=t−πω)=∫−πω∞dsϕAB(s+πω)eiωseiπ=−(∫0∞dsϕAB(s+πω)eiωs+∫−πω0dsϕAB(s+πω)eiωs)∴2I=∫0∞dt(ϕAB(t)−ϕAB(t+πω))eiωt−∫−πω0dtϕAB(t+πω)eiωt∴∣2I∣≤∣∫0∞dt(ϕAB(t)−ϕAB(t+πω))eiωt∣+∣∫−πω0dtϕAB(t+πω)eiωt∣≤∫0∞dt∣ϕAB(t)−ϕAB(t+πω)∣∣eiωt∣+∫−πω0dt∣ϕAB(t+πω)∣∣eiωt∣<∫0∞dt∣ϕAB(t)−ϕAB(t+πω)∣+∫−πω0dt∣ϕAB(t+πω)∣     (∵Im[ω]>0)\begin{aligned} I &\coloneqq \int^{\infty}_0 dt \phi_{AB(t)} e^{i\omega t} \\ &= \int^{\infty}_{- \frac{\pi}{\omega}} ds \phi_{AB(s + \frac{\pi}{\omega})} e^{i \omega (s + \frac{\pi}{\omega})} \ \ \ \ \ (s = t - \frac{\pi}{\omega}) \\ &= \int^{\infty}_{- \frac{\pi}{\omega}} ds \phi_{AB(s + \frac{\pi}{\omega})} e^{i \omega s} e^{i \pi} \\ &= - \left( \int^{\infty}_0 ds \phi_{AB(s + \frac{\pi}{\omega})} e^{i \omega s} + \int^0_{- \frac{\pi}{\omega}} ds \phi_{AB(s + \frac{\pi}{\omega})} e^{i \omega s} \right) \\ \therefore 2I &= \int^{\infty}_0 dt \left( \phi_{AB(t)} - \phi_{AB(t + \frac{\pi}{\omega})} \right) e^{i \omega t} - \int^{0}_{- \frac{\pi}{\omega}} dt \phi_{AB(t + \frac{\pi}{\omega})} e^{i \omega t} \\ \therefore \left| 2I \right| &\le \left| \int^{\infty}_0 dt \left( \phi_{AB(t)} - \phi_{AB(t + \frac{\pi}{\omega})} \right) e^{i \omega t} \right| + \left| \int^{0}_{- \frac{\pi}{\omega}} dt \phi_{AB(t + \frac{\pi}{\omega})} e^{i \omega t} \right| \\ &\le \int^{\infty}_0 dt \left| \phi_{AB(t)} - \phi_{AB(t + \frac{\pi}{\omega})} \right| \left| e^{i \omega t} \right| + \int^{0}_{- \frac{\pi}{\omega}} dt \left| \phi_{AB(t + \frac{\pi}{\omega})} \right| \left| e^{i \omega t} \right| \\ &\lt \int^{\infty}_0 dt \left| \phi_{AB(t)} - \phi_{AB(t + \frac{\pi}{\omega})} \right| + \int^{0}_{- \frac{\pi}{\omega}} dt \left| \phi_{AB(t + \frac{\pi}{\omega})} \right| \ \ \ \ \ (\because \text{Im}[\omega] > 0) \\ \end{aligned}

∫0∞dt∣ϕAB(t)−ϕAB(t+πω)∣=lim⁡τ→∞∫0τdt∣ϕAB(t)−ϕAB(t+πω)∣≤lim⁡τ→∞∫0τdt(∣ϕAB(t)∣+∣ϕAB(t+πω)∣)=2∫0∞dt∣ϕAB(t)∣     (⇒有界かつτについて単調増加)\begin{aligned} \int^{\infty}_0 dt \left| \phi_{AB(t)} - \phi_{AB(t + \frac{\pi}{\omega})} \right| &= \lim_{\tau \to \infty} \int^{\tau}_0 dt \left| \phi_{AB(t)} - \phi_{AB(t + \frac{\pi}{\omega})} \right| \\ &\le \lim_{\tau \to \infty} \int^{\tau}_0 dt \left( \left| \phi_{AB(t)} \right| + \left| \phi_{AB(t + \frac{\pi}{\omega})} \right| \right) \\ &= 2 \int^{\infty}_0 dt \left| \phi_{AB(t)} \right| \ \ \ \ \ (\Rightarrow \text{有界かつ}\tau\text{について単調増加}) \end{aligned}

∴lim⁡ω→∞∣I∣=lim⁡Δt→012∫0∞dt∣ϕAB(t+Δt)−ϕAB(t)∣=lim⁡Δt→012lim⁡Δt′→0∑i∣ϕAB(ti+Δt)−ϕAB(ti)∣Δt′=lim⁡Δt→012∑i∣ϕAB(ti+1)−ϕAB(ti)∣Δt     (Δt′=Δtの条件で極限をとる)≤lim⁡Δt→012∑i(Mi−mi)Δt=lim⁡Δt→012(S(Δt)−s(Δt))=0     (∵ϕAB(t)はリーマン可積分)\begin{aligned} \therefore \lim_{\omega \to \infty} \left| I \right| &= \lim_{\Delta t \to 0} \frac{1}{2} \int^{\infty}_0 dt \left| \phi_{AB(t + \Delta t)} - \phi_{AB(t)} \right| \\ &= \lim_{\Delta t \to 0} \frac{1}{2} \lim_{\Delta t' \to 0} \sum_{i} \left| \phi_{AB(t_i + \Delta t)} - \phi_{AB(t_i)} \right| \Delta t' \\ &= \lim_{\Delta t \to 0} \frac{1}{2} \sum_{i} \left| \phi_{AB(t_{i+1})} - \phi_{AB(t_i)} \right| \Delta t \ \ \ \ \ (\Delta t' = \Delta t \text{の条件で極限をとる})\\ &\le \lim_{\Delta t \to 0} \frac{1}{2} \sum_{i} ( M_i - m_i ) \Delta t \\ &= \lim_{\Delta t \to 0} \frac{1}{2} \left( S_{(\Delta t)} - s_{(\Delta t)} \right) \\ &= 0 \ \ \ \ \ (\because \phi_{AB(t)}\text{はリーマン可積分}) \end{aligned}

複素アドミッタンス

χAB(ω)=∫0∞ϕAB(t)eiωt\begin{aligned} \chi_{AB(\omega)} &= \int^{\infty}_0 \phi_{AB(t)} e^{i\omega t} \end{aligned}

∫badxϵx2+ϵ2f(x)=12i∫badx(1x−iϵ−1x+iϵ)f(x)=12i∫b−iϵa−iϵdzf(z+iϵ)z+∫a+iϵb+iϵdzf(z−iϵ)z=12i∮dzf(z)z+O(ϵ)→{πf(0)     (原点が積分経路に含まれる)0     (原点が積分経路に含まれない)\begin{aligned} \int^a_b dx \frac{\epsilon}{x^2 + \epsilon^2} f_{(x)} &= \frac{1}{2i} \int^a_b dx \left( \frac{1}{x - i\epsilon} - \frac{1}{x + i\epsilon} \right) f_{(x)} \\ &= \frac{1}{2i} \int^{a - i\epsilon}_{b - i\epsilon} dz \frac{f_{(z + i\epsilon)}}{z} + \int^{b + i\epsilon} _{a + i\epsilon} dz \frac{f_{(z - i\epsilon)}}{z} \\ &= \frac{1}{2i} \oint dz \frac{f_{(z)}}{z} + \text{O}_{(\epsilon)} \\ &\to \left\{ \begin{split} &\pi f_{(0)} \ \ \ \ \ (\text{原点が積分経路に含まれる}) \\ &0 \ \ \ \ \ (\text{原点が積分経路に含まれない}) \end{split} \right. \end{aligned}

∴lim⁡ϵ→0ϵx2+ϵ2=πδ(x)\therefore \lim_{\epsilon \to 0} \frac{\epsilon}{x^2 + \epsilon^2} = \pi \delta(x)

1x−iϵ=x+iϵ(x−iϵ)(x+iϵ)=xx2+ϵ2+iϵx2+ϵ2→Pv1x+iπδ(x)\begin{aligned} \frac{1}{x - i\epsilon} &= \frac{x + i\epsilon}{(x - i\epsilon)(x + i\epsilon)} \\ &= \frac{x}{x^2 + \epsilon^2} + i \frac{\epsilon}{x^2 + \epsilon^2} \\ &\to \text{Pv} \frac{1}{x} + i \pi \delta_{(x)} \end{aligned}

複素平面に拡張された複素アドミッタンス

12πi∫−∞∞dω′χAB(ω′)ω′−z={χAB(z)     (Im[z]>0)0     (Im[z]<0)\frac{1}{2\pi i} \int^{\infty}_{- \infty} d\omega' \frac{\chi_{AB(\omega')}}{\omega' - z} = \left\{ \begin{aligned} &\chi_{AB(z)} \ \ \ \ \ (\text{Im}[z] > 0) \\ &0 \ \ \ \ \ (\text{Im}[z] < 0) \\ \end{aligned} \right.

12πi∫∞∞dω′χAB(ω′)ω′−(ω+iϵ)\frac{1}{2\pi i} \int^{\infty}_{\infty} d\omega' \frac{\chi_{AB(\omega')}}{\omega' - (\omega + i \epsilon)}

12πi∫−∞∞dω′χAB(ω′)ω′−(ω+iϵ)=12πi∫∞∞dω′χAB(ω′)(ω′−ω)−iϵ→12πi∫−∞∞dω′(PvχAB(ω′)ω′−ω+iπχAB(ω′)δ(ω′−ω))=12πiPv∫−∞∞dω′χAB(ω′)ω′−ω+12χAB(ω)\begin{aligned} \frac{1}{2\pi i} \int^{\infty}_{-\infty} d\omega' \frac{\chi_{AB(\omega')}}{\omega' - (\omega + i \epsilon)} &= \frac{1}{2\pi i} \int^{\infty}_{\infty} d\omega' \frac{\chi_{AB(\omega')}}{(\omega' - \omega) - i \epsilon} \\ &\to \frac{1}{2 \pi i} \int^{\infty}_{-\infty} d\omega' \left( \text{Pv} \frac{\chi_{AB(\omega')}}{\omega' - \omega} + i \pi \chi_{AB(\omega')} \delta_{(\omega' - \omega)} \right) \\ &= \frac{1}{2 \pi i} \text{Pv} \int^{\infty}_{-\infty} d\omega' \frac{\chi_{AB(\omega')}}{\omega' - \omega} + \frac{1}{2} \chi_{AB(\omega)} \\ \end{aligned}

クラマース・クローニッヒ関係 ⋯(2.30)\cdots (2.30)

χAB(ω)′+iχAB(ω)′′=12πiPv∫−∞∞dω′χAB(ω′)′+iχAB(ω′)′′ω′−ω+12(χAB(ω)′+iχAB(ω)′′)∴χAB(ω)′+iχAB(ω)′′=1πPv∫−∞∞dω′−iχAB(ω′)′+χAB(ω′)′′ω′−ω\begin{aligned} \chi'_{AB(\omega)} + i\chi''_{AB(\omega)} &= \frac{1}{2 \pi i} \text{Pv} \int^{\infty}_{- \infty} d\omega' \frac{\chi'_{AB(\omega')} + i \chi''_{AB(\omega')}}{\omega' - \omega} + \frac{1}{2} \left( \chi'_{AB(\omega)} + i \chi''_{AB(\omega)} \right) \\ \therefore \chi'_{AB(\omega)} + i\chi''_{AB(\omega)} &= \frac{1}{\pi} \text{Pv} \int^{\infty}_{-\infty} d\omega' \frac{-i \chi'_{AB(\omega')} + \chi''_{AB(\omega')}}{\omega' - \omega} \end{aligned}

χAB(ω)′=1πPv∫−∞∞dω′χAB(ω′)′′ω′−ωχAB(ω)′′=−1πPv∫−∞∞dω′χAB(ω′)′ω′−ω\begin{aligned} \chi'_{AB(\omega)} &= \frac{1}{\pi} \text{Pv} \int^{\infty}_{-\infty} d\omega' \frac{\chi''_{AB(\omega')}}{\omega' - \omega} \\ \chi''_{AB(\omega)} &= - \frac{1}{\pi} \text{Pv} \int^{\infty}_{-\infty} d\omega' \frac{\chi'_{AB(\omega')}}{\omega' - \omega} \\ \end{aligned}

2.2.2 力学応答の揺動散逸定理 (P. 27)

§ 2.2.2.1 パワーロス (P. 27)

Htot=H−BF(t)H_{tot} = H - BF_{(t)}

タイプ 1

ΔA(t)=B(t)\Delta A_{(t)} = B_{(t)}

タイプ 2

ΔA(t)=B˙(t)\Delta A_{(t)} = \dot{B}_{(t)}

一周期あたりの仕事 W ⋯(2.31)\cdots(2.31)

W=∫02πωdtdF(t)dtTr(∂Htot∂Fρ(t))=−∫02πωdtdF(t)dt⟨B(t)⟩eq     (∵∂Htot∂F=−B)\begin{aligned} W &= \int^{\frac{2\pi}{\omega}}_{0} dt \frac{dF_{(t)}}{dt} \text{Tr}\left( \frac{\partial H_{tot}} {\partial F} \rho_{(t)} \right) \\ &= - \int^{\frac{2\pi}{\omega}}_{0} dt \frac{dF_{(t)}}{dt} \langle B_{(t)} \rangle_{eq} \ \ \ \ \ \left( \because \frac{\partial H_{tot}} {\partial F} = -B \right) \\ \end{aligned}

単位時間あたりの仕事PP ⋯(2.32)\cdots(2.32)

P(1)=−ω2π∫02πωdtdF(t)dt⟨B(t)⟩eq=−ω2π∫02πωdtdF(t)dt⟨ΔA(t)⟩eq     (∵ΔA(t)=B(t))=−ω2π∫02πωdt(−ω)F0sin⁡(ωt)F0Re[χAB(ω)e−iωt]=ω2F022πRe[χAB(ω)∫02πωdtsin⁡(ωt)e−iωt]=ω2F022πRe[χAB(ω)∫02πωdt(−i)sin⁡2(ωt)]=ωF022χAB(ω)′′\begin{aligned} P^{(1)} &= - \frac{\omega}{2 \pi} \int^{\frac{2\pi}{\omega}}_{0} dt \frac{dF_{(t)}}{dt} \langle B_{(t)} \rangle_{eq} \\ &= - \frac{\omega}{2 \pi} \int^{\frac{2\pi}{\omega}}_{0} dt \frac{dF_{(t)}}{dt} \langle \Delta A_{(t)} \rangle_{eq} \ \ \ \ \ \left(\because \Delta A_{(t)} = B_{(t)} \right) \\ &= - \frac{\omega}{2 \pi} \int^{\frac{2\pi}{\omega}}_{0} dt (-\omega) F_0 \sin{(\omega t)} F_0 \text{Re} \left[ \chi_{AB(\omega)} e^{-i\omega t} \right] \\ &= \frac{\omega^2 F_0^2}{2 \pi} \text{Re} \left[ \chi_{AB(\omega)} \int^{\frac{2\pi}{\omega}}_{0} dt \sin{(\omega t)} e^{-i\omega t} \right] \\ &= \frac{\omega^2 F_0^2}{2 \pi} \text{Re} \left[ \chi_{AB(\omega)} \int^{\frac{2\pi}{\omega}}_{0} dt (-i) \sin^2{(\omega t)} \right] \\ &= \frac{\omega F_0^2}{2} \chi''_{AB(\omega)} \end{aligned}

P(2)=−ω2π∫02πωdtdF(t)dt⟨B(t)⟩eq=−ω2π[F(t)⟨B(t)⟩eq]02πω+ω2π∫02πωdtF(t)ddt⟨B(t)⟩eq=ω2π∫02πωdtF(t)⟨B˙(t)⟩eq     (∵定常状態の周期性)=ω2π∫02πωdtF(t)⟨ΔA(t)⟩eq     (∵ΔA(t)=B˙(t))=ω2π∫02πωdtF0cos⁡(ωt)F0Re[χAB(ω)e−iωt]=ω2πF02Re[χAB(ω)∫02πωdtcos⁡(ωt)e−iωt]=ω2πF02Re[χAB(ω)∫02πωdtcos⁡2(ωt)]=F022χAB(ω)′\begin{aligned} P^{(2)} &= - \frac{\omega}{2 \pi} \int^{\frac{2\pi}{\omega}}_{0} dt \frac{dF_{(t)}}{dt} \langle B_{(t)} \rangle_{eq} \\ &= - \frac{\omega}{2 \pi} \left[ F_{(t)} \langle B_{(t)} \rangle_{eq} \right]^{\frac{2\pi}{\omega}}_{0} + \frac{\omega}{2 \pi} \int^{\frac{2\pi}{\omega}}_{0} dt F_{(t)} \frac{d}{dt} \langle B_{(t)} \rangle_{eq} \\ &= \frac{\omega}{2 \pi} \int^{\frac{2\pi}{\omega}}_{0} dt F_{(t)} \langle \dot{B}_{(t)} \rangle_{eq} \ \ \ \ \ \left( \because \text{定常状態の周期性} \right) \\ &= \frac{\omega}{2 \pi} \int^{\frac{2\pi}{\omega}}_{0} dt F_{(t)} \langle \Delta A_{(t)} \rangle_{eq} \ \ \ \ \ \left( \because \Delta A_{(t)} = \dot{B}_{(t)} \right) \\ &= \frac{\omega}{2 \pi} \int^{\frac{2\pi}{\omega}}_{0} dt F_0 \cos{(\omega t)} F_0 \text{Re} \left[ \chi_{AB(\omega)} e^{-i\omega t} \right] \\ &= \frac{\omega}{2 \pi} F_{0}^{2} \text{Re} \left[ \chi_{AB(\omega)} \int^{\frac{2\pi}{\omega}}_{0} dt \cos{(\omega t)} e^{-i\omega t} \right] \\ &= \frac{\omega}{2 \pi} F_{0}^{2} \text{Re} \left[ \chi_{AB(\omega)} \int^{\frac{2\pi}{\omega}}_{0} dt \cos^2{(\omega t)} \right] \\ &= \frac{F_0^2}{2} \chi'_{AB(\omega)} \end{aligned}

等温過程応答係数χABT\chi^{T}_{AB} ⋯(2.34)\cdots(2.34)

χABT=∂∂F0Tr[e−β(H−F0B)A]Tr[e−β(H−F0B)]∣F0→0=1Z{∂∂F0Tr[e−β(H−F0B)A]−⟨A⟩eq∂∂F0Tr[e−β(H−F0B)]}∣F0→0=1Z{Tr[∫01dye−βH(1−y)(βB)e−βHyA]−⟨A⟩eqTr[∫01dye−βH(1−y)(βB)e−βHy]}=1ZTr[∫01dye−βH(1−y)(βB)e−βHy(A−⟨A⟩eq)]=1ZTr[∫0βdue(uH−βH)Be−uH(A−⟨A⟩eq)]     (u≔βy)=∫0βdu1ZTr[e−βHeuHBe−uH(A−⟨A⟩eq)]=∫0βdu⟨euHBe−uH(A−⟨A⟩eq)⟩eq=∫0βdu⟨B(−iℏu)(A−⟨A⟩eq)⟩eq     (∵euHを時間発展演算子とみなした)=∫0βdu⟨B(A(iℏu)−⟨A⟩eq)⟩eq     (∵平衡状態の平均は時間に依存しない)=∫0βdu(⟨BA(iℏu)⟩eq−⟨B⟩eq⟨A⟩eq)=∫0βdu⟨ΔBΔA(iℏu)⟩eq     (ref.  Cov(xy)=E[xy]−E[x]E[y])\begin{aligned} \chi^{T}_{AB} &= \left. \frac{\partial}{\partial F_0} \frac{\text{Tr}\left[ e^{-\beta (H - F_0 B)} A \right]}{\text{Tr}\left[ e^{-\beta(H - F_0B)} \right]} \right|_{F_0 \to 0} \\ &= \left. \frac{1}{Z} \left\{ \frac{\partial}{\partial F_0} \text{Tr}\left[ e^{-\beta (H - F_0 B)} A \right] - \langle A \rangle_{eq} \frac{\partial}{\partial F_0} \text{Tr}\left[ e^{-\beta(H - F_0B)} \right] \right\} \right|_{F_0 \to 0} \\ &= \frac{1}{Z} \left\{ \text{Tr} \left[ \int^1_0 dy e^{-\beta H(1-y)} (\beta B) e^{-\beta H y}A \right] - \langle A \rangle_{eq} \text{Tr} \left[\int^1_0 dy e^{-\beta H(1-y)} (\beta B) e^{-\beta H y} \right] \right\} \\ &= \frac{1}{Z} \text{Tr} \left[\int^1_0 dy e^{-\beta H(1-y)} (\beta B) e^{-\beta H y} (A - \langle A \rangle_{eq}) \right] \\ &= \frac{1}{Z} \text{Tr} \left[\int^{\beta}_0 du e^{(u H-\beta H)} B e^{-u H} (A - \langle A \rangle_{eq}) \right] \ \ \ \ \ (u \coloneqq \beta y) \\ &= \int^{\beta}_0 du \frac{1}{Z} \text{Tr} \left[ e^{-\beta H} e^{u H} B e^{-u H} (A - \langle A \rangle_{eq}) \right] \\ &= \int^{\beta}_0 du \langle e^{u H} B e^{-u H} (A - \langle A \rangle_{eq}) \rangle_{eq} \\ &= \int^{\beta}_0 du \langle B_{(-i\hbar u)} (A - \langle A \rangle_{eq}) \rangle_{eq} \ \ \ \ \ \left( \because e^{u H}\text{を時間発展演算子とみなした} \right) \\ &= \int^{\beta}_0 du \langle B (A_{(i \hbar u)} - \langle A \rangle_{eq}) \rangle_{eq} \ \ \ \ \ \left( \because \text{平衡状態の平均は時間に依存しない} \right) \\ &= \int^{\beta}_0 du \left( \langle B A_{(i \hbar u)} \rangle_{eq} - \langle B \rangle_{eq} \langle A \rangle_{eq} \right) \\ &= \int^{\beta}_0 du \langle \Delta B \Delta A_{(i \hbar u)} \rangle_{eq} \ \ \ \ \ \left( \text{ref.} \ \ \text{Cov}(xy) = \text{E}[xy] - \text{E}[x]\text{E}[y] \right) \\ \end{aligned}

熱力学的応答関数と力学的応答関数の差 ⋯(2.35)\cdots(2.35)

lim⁡ω→0χAB(ω)=lim⁡ω→0∫0∞dtϕAB(t)e−iωt=lim⁡τ→∞∫0τdtϕAB(t)=lim⁡τ→∞∫0τdt∫0βdu⟨B˙(−iℏu)A(t)⟩eq     (∵(2.25))=−lim⁡τ→∞∫0τdt∫0βduddt⟨BA(t+iℏu)⟩eq     (∵(公式1))=−∫0βdulim⁡τ→∞[⟨BA(t+iℏu)⟩eq]0τ=∫0βdu(⟨BA(iℏu)⟩eq−lim⁡τ→∞⟨BA(τ+iℏu)⟩eq)\begin{aligned} \lim_{\omega \to 0} \chi_{AB(\omega)} &= \lim_{\omega \to 0} \int^{\infty}_0 dt \phi_{AB(t)} e^{-i \omega t} \\ &= \lim_{\tau \to \infty} \int^{\tau}_0 dt \phi_{AB(t)} \\ &= \lim_{\tau \to \infty} \int^{\tau}_0 dt \int^{\beta}_0 du \langle \dot{B}_{(-i \hbar u)} A_{(t)} \rangle_{eq} \ \ \ \ \ (\because \text{(2.25)}) \\ &= - \lim_{\tau \to \infty} \int^{\tau}_0 dt \int^{\beta}_0 du \frac{d}{dt} \langle B A_{(t + i \hbar u)} \rangle_{eq} \ \ \ \ \ (\because (公式1)) \\ &= - \int^{\beta}_0 du \lim_{\tau \to \infty} \left[ \langle B A_{(t + i \hbar u)} \rangle_{eq} \right]^{\tau}_0 \\ &= \int^{\beta}_0 du \left( \langle B A_{(i \hbar u)} \rangle_{eq} - \lim_{\tau \to \infty} \langle B A_{(\tau + i \hbar u)} \rangle_{eq} \right) \end{aligned}

χABT−χAB=∫0βdu⟨ΔBΔA(iℏu)⟩−∫0βdu(⟨BA(iℏu)⟩eq−lim⁡τ→∞⟨BA(τ+iℏu)⟩eq)=∫0βdu(⟨BA(iℏu)⟩eq−⟨B⟩eq⟨A⟩eq)−∫0βdu(⟨BA(iℏu)⟩eq−lim⁡τ→∞⟨BA(τ+iℏu)⟩eq)=∫0βdu(lim⁡τ→∞⟨BA(τ+iℏu)⟩eq−⟨B⟩eq⟨A⟩eq)=lim⁡τ→∞∫0βdu(⟨BA(τ+iℏu)⟩eq−⟨B⟩eq⟨A⟩eq)=lim⁡τ→∞∫0βdu⟨ΔBΔA(τ+iℏu)⟩eq\begin{aligned} \chi^{T}_{AB} - \chi_{AB} &= \int^{\beta}_0 du \langle \Delta B \Delta A_{(i \hbar u)} \rangle - \int^{\beta}_0 du \left( \langle B A_{(i \hbar u)} \rangle_{eq} - \lim_{\tau \to \infty} \langle B A_{(\tau + i \hbar u)} \rangle_{eq} \right) \\ &= \int^{\beta}_0 du \left( \langle B A_{(i \hbar u)} \rangle_{eq} - \langle B \rangle_{eq} \langle A \rangle_{eq} \right) - \int^{\beta}_0 du \left( \langle B A_{(i \hbar u)} \rangle_{eq} - \lim_{\tau \to \infty} \langle B A_{(\tau + i \hbar u)} \rangle_{eq} \right) \\ &= \int^{\beta}_0 du \left( \lim_{\tau \to \infty} \langle B A_{(\tau + i \hbar u)} \rangle_{eq} - \langle B \rangle_{eq} \langle A \rangle_{eq} \right) \\ &= \lim_{\tau \to \infty} \int^{\beta}_0 du \left( \langle B A_{(\tau + i \hbar u)} \rangle_{eq} - \langle B \rangle_{eq} \langle A \rangle_{eq} \right) \\ &= \lim_{\tau \to \infty} \int^{\beta}_0 du \langle \Delta B \Delta A_{(\tau + i \hbar u)} \rangle_{eq} \end{aligned}

§ 2.2.2.2 揺動散逸定理 (P. 29)

時間反転

⟨[A(t),B]⟩eq=Tr[π[A(t),B]]=Tr[(Tπ[A(t),B]T†)†]=Tr[[B†,A(−t)†]π†]=−Tr[[A(−t),B]π]=−⟨[A(−t),B]⟩eq\begin{aligned} \langle [A_{(t)}, B] \rangle_{eq} &= \text{Tr} \left[ \pi [A_{(t)}, B] \right] \\ &= \text{Tr} \left[ \left( T \pi [A_{(t)}, B] T^{\dagger} \right)^{\dagger} \right] \\ &= \text{Tr} \left[ [B^{\dagger}, A^{\dagger}_{(-t)}] \pi^{\dagger} \right] \\ &= - \text{Tr} \left[ [A_{(-t)}, B] \pi \right] \\ &= - \langle [A_{(-t)}, B] \rangle_{eq} \\ \end{aligned}

⟨[A(t),B]⟩eq=⟨[B(t),B]⟩eq=⟨[B,B(−t)]⟩eq=−⟨[B(−t),B]⟩eq=−⟨[A(−t),B]⟩eq\begin{aligned} \langle [A_{(t)}, B] \rangle_{eq} &= \langle [B_{(t)}, B] \rangle_{eq} \\ &= \langle [B, B_{(-t)}] \rangle_{eq} \\ &= - \langle [B_{(-t)}, B] \rangle_{eq} \\ &= - \langle [A_{(-t)}, B] \rangle_{eq} \\ \end{aligned}

⟨[A(t),B]⟩eq=⟨[B˙(t),B]⟩eq=ddt⟨[B(t),B]⟩eq=ddt⟨[B,B(−t)]⟩eq=−ddt⟨[B(−t),B]⟩eq=⟨[B˙(−t),B]⟩eq=⟨[A(−t),B]⟩eq\begin{aligned} \langle [A_{(t)}, B] \rangle_{eq} &= \langle [\dot{B}_{(t)}, B] \rangle_{eq} \\ &= \frac{d}{dt} \langle [B_{(t)}, B] \rangle_{eq} \\ &= \frac{d}{dt} \langle [B, B_{(-t)}] \rangle_{eq} \\ &= - \frac{d}{dt} \langle [B_{(-t)}, B] \rangle_{eq} \\ &= \langle [\dot{B}_{(-t)}, B] \rangle_{eq} \\ &= \langle [A_{(-t)}, B] \rangle_{eq} \\ \end{aligned}

応答関数ϕAB\phi_{AB}

ϕAB(ω)=∫∞∞dtiℏ⟨[A(t),B]⟩eqeiωt=∫0∞dtiℏ⟨[A(t),B]⟩eq(eiωt−e−iωt)=2i∫0∞dtiℏ⟨[A(t),B]⟩eqIm[eiωt]=2iIm[∫0∞dtiℏ⟨[A(t),B]⟩eqeiωt]=2iIm[χAB(ω)]=2χAB(ω)′\begin{aligned} \phi_{AB(\omega)} &= \int^{\infty}_{\infty} dt \frac{i}{\hbar} \langle [A_{(t)}, B] \rangle_{eq} e^{i\omega t} \\ &= \int^{\infty}_0 dt \frac{i}{\hbar} \langle [A_{(t)}, B] \rangle_{eq} (e^{i \omega t} - e^{-i \omega t}) \\ &= 2i \int^{\infty}_0 dt \frac{i}{\hbar} \langle [A_{(t)}, B] \rangle_{eq} \text{Im}[e^{i \omega t}] \\ &= 2i \text{Im} \left[ \int^{\infty}_0 dt \frac{i}{\hbar} \langle [A_{(t)}, B] \rangle_{eq} e^{i \omega t} \right] \\ &= 2i \text{Im} [\chi_{AB(\omega)}] \\ &= 2 \chi'_{AB(\omega)} \end{aligned}

ϕAB(ω)=∫∞∞dtiℏ⟨[A(t),B]⟩eqeiωt=∫0∞dtiℏ⟨[A(t),B]⟩eq(eiωt+e−iωt)=2∫0∞dtiℏ⟨[A(t),B]⟩eqRe[eiωt]=2Re[∫0∞dtiℏ⟨[A(t),B]⟩eqeiωt]=2Re[χAB(ω)]=2χAB(ω)′′\begin{aligned} \phi_{AB(\omega)} &= \int^{\infty}_{\infty} dt \frac{i}{\hbar} \langle [A_{(t)}, B] \rangle_{eq} e^{i\omega t} \\ &= \int^{\infty}_0 dt \frac{i}{\hbar} \langle [A_{(t)}, B] \rangle_{eq} (e^{i \omega t} + e^{-i \omega t}) \\ &= 2 \int^{\infty}_0 dt \frac{i}{\hbar} \langle [A_{(t)}, B] \rangle_{eq} \text{Re}[e^{i \omega t}] \\ &= 2 \text{Re} \left[ \int^{\infty}_0 dt \frac{i}{\hbar} \langle [A_{(t)}, B] \rangle_{eq} e^{i \omega t} \right] \\ &= 2 \text{Re} [\chi_{AB(\omega)}] \\ &= 2 \chi''_{AB(\omega)} \end{aligned}

KMS 条件

⟨AB(t+iℏβ)⟩eq=Tr[πAe−βHB(t)eβH]=1ZTr[e−βHAe−βHB(t)eβH]=1ZTr[Ae−βHB(t)eβHe−βH]=1ZTr[Ae−βHB(t)]=1ZTr[e−βHB(t)A]=⟨B(t)A⟩eq\begin{aligned} \langle AB_{(t + i\hbar\beta)} \rangle_{eq} &= \text{Tr}[\pi A e^{-\beta H} B_{(t)} e^{\beta H}] \\ &= \frac{1}{Z} \text{Tr} [e^{-\beta H} A e^{-\beta H} B_{(t)} e^{\beta H}] \\ &= \frac{1}{Z} \text{Tr} [A e^{-\beta H} B_{(t)} e^{\beta H} e^{-\beta H}] \\ &= \frac{1}{Z} \text{Tr} [A e^{-\beta H} B_{(t)}] \\ &= \frac{1}{Z} \text{Tr} [e^{-\beta H} B_{(t)} A] \\ &= \langle B_{(t)} A \rangle_{eq} \end{aligned}

⟨ΔBΔA(t+iℏβ)⟩eq=⟨ΔA(t)ΔB⟩eq\langle \Delta B \Delta A_{(t + i\hbar\beta)} \rangle_{eq} = \langle \Delta A_{(t)} \Delta B \rangle_{eq}

∫Cdz⟨ΔBΔA(z)⟩eqeiωz=0∴∫−∞∞dt⟨ΔBΔA(t)⟩eqeiωt=∫−∞∞dt⟨ΔBΔA(t+iℏβ)⟩eqeiω(t+iℏβ)=e−ℏβω∫−∞∞dt⟨ΔBΔA(t+iℏβ)⟩eqeiωt=e−ℏβω∫−∞∞dt⟨ΔA(t)ΔB⟩eqeiωt\begin{aligned} \int_{C} dz \langle \Delta B \Delta A_{(z)} \rangle_{eq} e^{i \omega z} &= 0 \\ \therefore \int^{\infty}_{-\infty} dt \langle \Delta B \Delta A_{(t)} \rangle_{eq} e^{i \omega t} &= \int^{\infty}_{-\infty} dt \langle \Delta B \Delta A_{(t + i \hbar \beta)} \rangle_{eq} e^{i \omega (t + i \hbar \beta)} \\ &= e^{- \hbar \beta \omega}\int^{\infty}_{-\infty} dt \langle \Delta B \Delta A_{(t + i \hbar \beta)} \rangle_{eq} e^{i \omega t} \\ &= e^{- \hbar \beta \omega} \int^{\infty}_{-\infty} dt \langle \Delta A_{(t)} \Delta B \rangle_{eq} e^{i \omega t} \\ \end{aligned}

応答関数ϕAB\phi_{AB}

[A(t),B]=[ΔA(t)+⟨A⟩eq,ΔB+⟨B⟩eq]=[ΔA(t),ΔB]\begin{aligned} [ A_{(t)}, B ] &= [ \Delta A_{(t)} + \langle A \rangle_{eq}, \Delta B + \langle B \rangle_{eq}] \\ &= [ \Delta A_{(t)}, \Delta B ] \end{aligned}

ϕAB=iℏ∫−∞∞dt⟨[A(t),B]⟩eqeiωt=iℏ∫−∞∞dt⟨[ΔA(t),ΔB]⟩eqeiωt=iℏ∫−∞∞dt(⟨ΔA(t)ΔB⟩eq−⟨ΔBΔA(t)⟩eq)eiωt=iℏ∫−∞∞dt(⟨ΔA(t)ΔB⟩eq−e−ℏβω⟨ΔA(t)ΔB⟩eq)eiωt=iℏ(1−e−ℏβω)∫−∞∞dt⟨ΔA(t)ΔB⟩eqeiωt\begin{aligned} \phi_{AB} &= \frac{i}{\hbar} \int^{\infty}_{-\infty} dt \langle [A_{(t)}, B] \rangle_{eq} e^{i \omega t} \\ &= \frac{i}{\hbar} \int^{\infty}_{- \infty} dt \langle [ \Delta A_{(t)}, \Delta B ] \rangle_{eq} e^{i \omega t} \\ &= \frac{i}{\hbar} \int^{\infty}_{- \infty} dt \left( \langle \Delta A_{(t)}\Delta B \rangle_{eq} - \langle \Delta B \Delta A_{(t)} \rangle_{eq} \right) e^{i \omega t} \\ &= \frac{i}{\hbar} \int^{\infty}_{- \infty} dt \left( \langle \Delta A_{(t)}\Delta B \rangle_{eq} - e^{-\hbar\beta\omega} \langle \Delta A_{(t)} \Delta B \rangle_{eq} \right) e^{i \omega t} \\ &= \frac{i}{\hbar} (1 - e^{-\hbar\beta\omega}) \int^{\infty}_{- \infty} dt \langle \Delta A_{(t)}\Delta B \rangle_{eq} e^{i \omega t} \\ \end{aligned}

ϕAB=iℏ(1−e−ℏβω)∫−∞∞dt⟨ΔA(t)ΔB⟩eqeiωt=(iℏ1−e−ℏβω)∫−∞∞dteℏβω⟨ΔBΔA(t)⟩eqeiωt=iℏ(eℏβω−1)∫−∞∞dt⟨ΔBΔA(t)⟩eqeiωt\begin{aligned} \phi_{AB} &= \frac{i}{\hbar} (1 - e^{-\hbar\beta\omega}) \int^{\infty}_{- \infty} dt \langle \Delta A_{(t)}\Delta B \rangle_{eq} e^{i \omega t} \\ &= (\frac{i}{\hbar} 1 - e^{-\hbar\beta\omega}) \int^{\infty}_{- \infty} dt e^{\hbar\beta\omega} \langle \Delta B \Delta A_{(t)} \rangle_{eq} e^{i \omega t} \\ &= \frac{i}{\hbar} (e^{\hbar\beta\omega} - 1) \int^{\infty}_{- \infty} dt \langle \Delta B \Delta A_{(t)} \rangle_{eq} e^{i \omega t} \\ \end{aligned}

CAB=∫−∞∞dt12[⟨ΔA(t)ΔB⟩eq+⟨ΔBΔA(t)⟩eq]eiωt=12[∫−∞∞dt⟨ΔA(t)ΔB⟩eqeiωt+∫−∞∞dt⟨ΔBΔA(t)⟩eqeiωt]=12[1iℏ(1−e−ℏβω)ϕAB+1iℏ(eℏβω−1)ϕAB]=ℏ2ie12ℏβω+e−12ℏβωe12ℏβω−e−12ℏβωϕAB=ℏω2iωcoth⁡(ℏβω2)ϕAB∴ϕAB=iω(ℏω2)coth⁡(βℏω2)CAB\begin{aligned} C_{AB} &= \int^{\infty}_{-\infty} dt \frac{1}{2} \left[ \langle \Delta A_{(t)} \Delta B \rangle_{eq} + \langle \Delta B \Delta A_{(t)} \rangle_{eq} \right] e^{i\omega t} \\ &= \frac{1}{2} \left[ \int^{\infty}_{-\infty} dt \langle \Delta A_{(t)} \Delta B \rangle_{eq} e^{i\omega t} + \int^{\infty}_{-\infty} dt \langle \Delta B \Delta A_{(t)} \rangle_{eq} e^{i\omega t} \right] \\ &= \frac{1}{2} \left[ \frac{1}{\frac{i}{\hbar} (1 - e^{-\hbar\beta\omega})} \phi_{AB} + \frac{1}{\frac{i}{\hbar} (e^{\hbar\beta\omega} - 1)} \phi_{AB} \right] \\ &= \frac{\hbar}{2i} \frac{e^{\frac12 \hbar\beta\omega} + e^{-\frac12 \hbar\beta\omega}}{e^{\frac12 \hbar\beta\omega} - e^{-\frac12 \hbar\beta\omega}} \phi_{AB} \\ &= \frac{\frac{\hbar\omega}{2}}{i\omega} \coth{\left( \frac{\hbar\beta\omega}{2} \right)} \phi_{AB} \\ \therefore \phi_{AB} &= \frac{i\omega}{\left( \frac{\hbar\omega}{2} \right) \coth{\left( \frac{\beta\hbar\omega}{2} \right)}} C_{AB} \end{aligned}

散逸とゆらぎの関係

χAB′′=12iϕAB=1ℏcoth⁡(βℏω2)CAB\begin{aligned} \chi''_{AB} &= \frac{1}{2i} \phi_{AB} \\ &= \frac{1}{\hbar \coth{\left( \frac{\beta\hbar\omega}{2} \right)}} C_{AB} \end{aligned}

CAB˙=∫−∞∞dt12[⟨ΔA(t)ΔB˙⟩eq+⟨ΔB˙ΔA(t)⟩eq]eiωt=−∫−∞∞dt12(ddt[⟨ΔA(t)ΔB⟩eq+⟨ΔBΔA(t)⟩eq])eiωt     (∵(公式1))=∫−∞∞dt12[⟨ΔA(t)ΔB⟩eq+⟨ΔBΔA(t)⟩eq]ddteiωt     (∵混合性)=iω∫−∞∞dt12[⟨ΔA(t)ΔB⟩eq+⟨ΔBΔA(t)⟩eq]eiωt=iωCAB\begin{aligned} C_{A\dot{B}} &= \int^{\infty}_{-\infty} dt \frac12 \left[ \langle \Delta A_{(t)} \Delta \dot{B} \rangle_{eq} + \langle \Delta \dot{B} \Delta A_{(t)} \rangle_{eq} \right] e^{i\omega t} \\ &= - \int^{\infty}_{-\infty} dt \frac12 \left( \frac{d}{dt} \left[ \langle \Delta A_{(t)} \Delta B \rangle_{eq} + \langle \Delta B \Delta A_{(t)} \rangle_{eq} \right] \right) e^{i\omega t} \ \ \ \ \ (\because (\text{公式1})) \\ &= \int^{\infty}_{-\infty} dt \frac12 \left[ \langle \Delta A_{(t)} \Delta B \rangle_{eq} + \langle \Delta B \Delta A_{(t)} \rangle_{eq} \right] \frac{d}{dt} e^{i\omega t} \ \ \ \ \ (\because \text{混合性}) \\ &= i \omega \int^{\infty}_{-\infty} dt \frac12 \left[ \langle \Delta A_{(t)} \Delta B \rangle_{eq} + \langle \Delta B \Delta A_{(t)} \rangle_{eq} \right] e^{i\omega t} \\ &= i \omega C_{AB} \\ \end{aligned}

χAB′=12ϕAB=iωℏωcoth⁡(βℏω2)CAB=1ℏωcoth⁡(βℏω2)CAB˙\begin{aligned} \chi'_{AB} &= \frac12 \phi_{AB} \\ &= \frac{i\omega}{\hbar\omega \coth{\left( \frac{\beta\hbar\omega}{2} \right)}} C_{AB} \\ &= \frac{1}{\hbar\omega \coth{\left( \frac{\beta\hbar\omega}{2} \right)}} C_{A\dot{B}} \\ \end{aligned}

タイプ 2: ω→0\omega \to 0の極限

χAB′=1ℏωcoth⁡(βℏω2)CAB˙→β2βℏω2βℏω2∫−∞∞dtCAB˙(to)     (∵1coth⁡(x)=tanh⁡(x)=x+O(x3))=β2∫−∞∞dtCAB˙(t)\begin{aligned} \chi'_{AB} &= \frac{1}{\hbar\omega \coth{\left( \frac{\beta\hbar\omega}{2} \right)}} C_{A\dot{B}} \\ &\to \frac{\frac{\beta}{2} \frac{\beta\hbar\omega}{2}}{\frac{\beta\hbar\omega}{2}} \int^{\infty}_{-\infty} dt C_{A\dot{B}(to)} \ \ \ \ \ \left( \because \frac{1}{\coth{(x)}} = \tanh{(x)} = x + \text{O}(x^3) \right) \\ &= \frac{\beta}{2} \int^{\infty}_{-\infty} dt C_{A\dot{B}(t)} \end{aligned}

2.2.3 分布応答の線形応答と揺動散逸定理 (P. 31)

Htot=Hs+∑kHk+HksH_{tot} = H_s + \sum_{k} H_k + H_{ks}

分布関数 ρ0\rho_0

σu≔e−u(G+X)σˉu≔euGσu\begin{aligned} \sigma_u &\coloneqq e^{-u ( G + X )} \\ \bar{\sigma}_u &\coloneqq e^{uG}\sigma_u \end{aligned}

∂∂uσˉu=euGGσu−euG(G+X)σu=−euGXσu=−euGXe−uGσˉu=−X(−iℏu)σˉu∴σˉβ=e−∫0βduX(−iℏu)∴σβ=e−uGe−∫0βduX(−iℏu)\begin{aligned} \frac{\partial}{\partial u} \bar{\sigma}_u &= e^{uG} G \sigma_u - e^{uG} (G + X) \sigma_u \\ &= - e^{uG} X \sigma_u \\ &= - e^{uG} X e^{-uG} \bar{\sigma}_u \\ &= - X_{(-i\hbar u)} \bar{\sigma}_u \\ \therefore \bar{\sigma}_\beta &= e^{-\int^{\beta}_0 du X_{(-i \hbar u)}} \\ \therefore \sigma_\beta &= e^{-uG} e^{-\int^{\beta}_0 du X_{(-i \hbar u)}} \\ \end{aligned}

σβ=e−βGe−∫0βduX(−iℏu)=e−βG(1−∫0βduX(−iℏu))+O(X2)σβ=e−∫0βduX(iℏu)e−uG     (∵σβはエルミート)=(1−∫0βduX(iℏu))e−βG+O(X2)∴σβ≃e−βG−∫0βdu12[e−βGX(−iℏu)+X(iℏu)e−βG]\begin{aligned} \sigma_\beta &= e^{-\beta G} e^{-\int^{\beta}_0 du X_{(-i \hbar u)}} \\ &= e^{-\beta G} \left( 1 - \int^{\beta}_0 du X_{(-i \hbar u)} \right) + \text{O}_{(X^2)} \\ \sigma_\beta &= e^{-\int^{\beta}_0 du X_{(i \hbar u)}} e^{-uG} \ \ \ \ \ (\because \sigma_\beta \text{はエルミート}) \\ &= \left( 1 - \int^{\beta}_0 du X_{(i \hbar u)} \right) e^{-\beta G} + \text{O}_{(X^2)} \\ \therefore \sigma_\beta &\simeq e^{-\beta G} - \int^{\beta}_0 du \frac12 \left[ e^{-\beta G} X_{(-i \hbar u)} + X_{(i \hbar u)} e^{-\beta G} \right] \end{aligned}

Tr[e−u(G+X)]≃Tr[e−βG−∫0βdu12[e−βGX(−iℏu)+X(iℏu)e−βG]]=Tr[e−βG]−∫0βdu12(Tr[e−βGX(−iℏu)]+Tr[X(iℏu)e−βG])=Z−∫0βdu12(Z⟨X(−iℏu)⟩eq+Z⟨X(iℏu)⟩eq)=Z(1−∫0βdu12(⟨X(−iℏu)⟩eq+⟨X(iℏu)⟩eq))\begin{aligned} \text{Tr} \left[ e^{-u (G+X)} \right] &\simeq \text{Tr}\left[ e^{-\beta G} - \int^{\beta}_0 du \frac12 \left[ e^{-\beta G} X_{(-i \hbar u)} + X_{(i \hbar u)} e^{-\beta G} \right] \right] \\ &= \text{Tr}\left[ e^{-\beta G} \right] - \int^{\beta}_0 du \frac12 \left( \text{Tr}\left[ e^{-\beta G} X_{(-i \hbar u)} \right] + \text{Tr}\left[ X_{(i \hbar u)} e^{-\beta G} \right] \right) \\ &= Z - \int^{\beta}_0 du \frac{1}{2} ( Z \langle X_{(-i \hbar u)} \rangle_{eq} + Z \langle X_{(i \hbar u)} \rangle_{eq}) \\ &= Z \left( 1 - \int^{\beta}_0 du \frac{1}{2} ( \langle X_{(-i \hbar u)} \rangle_{eq} + \langle X_{(i \hbar u)} \rangle_{eq}) \right) \end{aligned}

11−x=1+x+O(x2)\frac{1}{1-x} = 1 + x + \text{O}_{(x^2)}

ρ0=e−β(G+X)Tre−β(G+X)≃(e−βG−∫0βdu12[e−βGX(−iℏu)+X(iℏu)e−βG]) 1Z(1+∫0βdu12(⟨X(−iℏu)⟩eq+⟨X(iℏu)⟩eq))=(π−∫0βdu12[πX(−iℏu)+X(iℏu)π])(1+∫0βdu12(⟨X(−iℏu)⟩eq+⟨X(iℏu)⟩eq))≃π−∫0βdu12[πX(−iℏu)+X(iℏu)π]+∫0βdu12(π⟨X(−iℏu)⟩eq+⟨X(iℏu)⟩eqπ)=π−∫0βdu12[πΔX(−iℏu)+ΔX(iℏu)π]\begin{aligned} \rho_0 &= \frac{e^{-\beta (G + X)}}{\text{Tr} e^{-\beta (G + X)}} \\ &\simeq \left( e^{-\beta G} - \int^{\beta}_0 du \frac12 \left[ e^{-\beta G} X_{(-i \hbar u)} + X_{(i \hbar u)} e^{-\beta G} \right] \right) \frac{1}{Z} \left( 1 + \int^{\beta}_0 du \frac{1}{2} ( \langle X_{(-i \hbar u)} \rangle_{eq} + \langle X_{(i \hbar u)} \rangle_{eq}) \right) \\ &= \left( \pi - \int^{\beta}_0 du \frac12 \left[ \pi X_{(-i \hbar u)} + X_{(i \hbar u)} \pi \right] \right) \left( 1 + \int^{\beta}_0 du \frac{1}{2} ( \langle X_{(-i \hbar u)} \rangle_{eq} + \langle X_{(i \hbar u)} \rangle_{eq}) \right) \\ &\simeq \pi - \int^{\beta}_0 du \frac12 \left[ \pi X_{(-i \hbar u)} + X_{(i \hbar u)} \pi \right] + \int^{\beta}_0 du \frac{1}{2} ( \pi \langle X_{(-i \hbar u)} \rangle_{eq} + \langle X_{(i \hbar u)} \rangle_{eq} \pi) \\ &= \pi - \int^{\beta}_0 du \frac12 \left[ \pi \Delta X_{(-i \hbar u)} + \Delta X_{(i \hbar u)} \pi \right] \end{aligned}

演算子の時間発展 ⋯(2.55)\cdots (2.55)

⟨A(τ)⟩eq=Tr[ρ0U†AU]=Tr[AUρ0U†]=Tr[AUπU†]−∫0βdu12(Tr[AUπΔX(−iℏu)U†]+Tr[AUΔX(iℏu)πU†])=Tr[πA]−∫0βdu12(Tr[πΔX(−iℏu−τ)A]+Tr[πAΔX(iℏu−τ)])=⟨A⟩eq−∫0βdu12[⟨ΔX(−iℏu−τ)A⟩eq+⟨AΔX(iℏu−τ)⟩eq]=⟨A⟩eq+∫0βdu∫−τ0dt12[⟨ΔX˙(−iℏu+t)⟩eq+⟨AΔX˙(iℏu+t)⟩eq]−∫0βdu12(⟨ΔX(−iℏu)A⟩eq+⟨AΔX(iℏu)⟩eq)\begin{aligned} \langle A_{(\tau)} \rangle_{eq} &= \text{Tr}[ \rho_0 U^{\dagger} A U] \\ &= \text{Tr}[A U \rho_0 U^{\dagger}] \\ &= \text{Tr}[AU\pi U^{\dagger}] - \int^{\beta}_0 du \frac12 \left( \text{Tr}[AU\pi\Delta X_{(-i \hbar u)}U^{\dagger}] + \text{Tr}[AU\Delta X_{(i \hbar u)} \pi U^{\dagger}] \right) \\ &= \text{Tr}[\pi A] - \int^{\beta}_0 du \frac12 \left( \text{Tr}[\pi\Delta X_{(-i \hbar u - \tau)} A] + \text{Tr}[\pi A\Delta X_{(i \hbar u - \tau)}] \right) \\ &= \langle A \rangle_{eq} - \int^{\beta}_0 du \frac12 \left[ \langle \Delta X_{(-i \hbar u - \tau)} A \rangle_{eq} + \langle A\Delta X_{(i \hbar u - \tau)} \rangle_{eq} \right] \\ &= \langle A \rangle_{eq} + \int^{\beta}_0 du \int^0_{-\tau} dt \frac12 \left[ \langle \Delta \dot{X}_{(-i \hbar u + t)} \rangle_{eq} + \langle A \Delta \dot{X}_{(i \hbar u + t)} \rangle_{eq} \right] - \int^{\beta}_0 du \frac12 \left( \langle \Delta X_{(-i \hbar u)} A \rangle_{eq} + \langle A\Delta X_{(i \hbar u)} \rangle_{eq} \right) \end{aligned}